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Worked solution

Separate signed attack advantage from damage clipped at zero

A board game uses eight-sided dice. The attack die has five faces marked 0, two marked 1 and one marked 2. After the attack throw, the defender throws a defence die with two faces marked 0, four marked 1 and two marked 2. Damage is attack score minus defence score, except that a negative difference gives zero damage.

Let A,D denote the attack and defence scores.

(i) Write down their distributions. Hence find the expectation and variance of A−DA-D. marks

X is the damage in one round.

(ii) Find its distribution, expectation and variance. marks

(iii) Explain why E(X)>0E(X)>0 although E(A)<E(D)E(A)<E(D). mark

[2017/RVHS/II/7]

CJC · Review problems · Question 11

Answer

E(A−D)=−1/2E(A-D)=-1/2, variance 1; X distribution above; E(X)=3/16E(X)=3/16, variance 55/256; positive damage has positive probability.

(i) Retain both source distributions and model

The source calls these dice “tetrahedral (8-sided)”; tetrahedra have four faces. The eight stated face counts determine this model. The source assumes equally likely physical faces and independent dice.

aP(A=a)P(A=a)
05/85/8
12/82/8
21/81/8
dP(D=d)P(D=d)
02/82/8
14/84/8
22/82/8
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(i) Retain the mean calculations

E(A)=0(5/8)+1(2/8)+2(1/8)=1/2E(A)=0(5/8)+1(2/8)+2(1/8)=1/2. E(D)=0(2/8)+1(4/8)+2(2/8)=1E(D)=0(2/8)+1(4/8)+2(2/8)=1. Thus E(A−D)=E(A)−E(D)=−1/2E(A-D)=E(A)-E(D)=-1/2.

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(i) Retain every second-moment and variance term

E(A2)=02(5/8)+12(2/8)+22(1/8)=3/4E(A^2)=0^2(5/8)+1^2(2/8)+2^2(1/8)=3/4 and E(D2)=02(2/8)+12(4/8)+22(2/8)=3/2E(D^2)=0^2(2/8)+1^2(4/8)+2^2(2/8)=3/2. Hence Var⁡(A)=3/4−(1/2)2=1/2\operatorname{Var}(A)=3/4-(1/2)^2=1/2 and Var⁡(D)=3/2−12=1/2\operatorname{Var}(D)=3/2-1^2=1/2. Independence gives Var⁡(A−D)=Var⁡(A)+Var⁡(D)=1/2+1/2=1\operatorname{Var}(A-D)=\operatorname{Var}(A)+\operatorname{Var}(D)=1/2+1/2=1.

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(ii) Retain every source damage case and probability

X=max⁡(A−D,0)X=\max(A-D,0).

xSource probability decompositionP(X=x)P(X=x)
0P(A=0)+P(A=1)P(D≥1)+P(A=2)P(D=2)P(A=0)+P(A=1)P(D\ge1)+P(A=2)P(D=2)27/3227/32
1P(A=1)P(D=0)+P(A=2)P(D=1)P(A=1)P(D=0)+P(A=2)P(D=1)1/81/8
2P(A=2)P(D=0)P(A=2)P(D=0)1/321/32
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(ii) Damage moments

E(X)=0(27/32)+1(1/8)+2(1/32)=3/16E(X)=0(27/32)+1(1/8)+2(1/32)=3/16. E(X2)=02(27/32)+12(1/8)+22(1/32)=1/4E(X^2)=0^2(27/32)+1^2(1/8)+2^2(1/32)=1/4. Hence Var⁡(X)=1/4−(3/16)2=55/256\operatorname{Var}(X)=1/4-(3/16)^2=55/256.

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(iii) Explain the strict positivity precisely

As the source explains, negative A−D outcomes become zero damage rather than cancelling positive damage. Thus X is nonnegative. For strict positivity, also note that positive damage has nonzero probability: P(X>0)=1/8+1/32=5/32>0P(X>0)=1/8+1/32=5/32>0. Therefore E(X)=3/16>0E(X)=3/16>0, even though the signed mean difference is −1/2. Clipping alone guarantees a nonnegative mean; here the positive cases make it strictly positive.

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Common mistakes

The nonnegative damage variable is not A−D. Strictly positive expectation also needs a positive-probability positive outcome.

Solution notes

All asks/marks, both original distributions, every mean/second-moment/variance term, all damage-case decompositions and source explanation retained. Eight-sided/tetrahedral source wording explicitly qualified; strict positivity justification completed using retained positive cases. Native original unchanged.

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